googologywikiaorg-20200223-history
User blog:Kyodaisuu/Mashimo scale
Mashimo scale is defined as \= floor(M^{-1}(x))\ where \(M(x)\) is Mashimo functionBlog post in Japanese Googology Wikija:マシモスケール. Therefore, \(M(n) \le x < M(n+1)\). For example, Mashimo scale of \(x=10^{34}\) is calculated as \(n = floor(M^{-1}(10^{34})) = floor(3.4) = 3\). Therefore, \(10^{34}\) is a large number of Mashimo scale 3. Here is a list of googological numbers classified with Mashimo scale. * \(\uparrow\) is Arrow notation * \(\rightarrow\) is Chained arrow notation * \(A\) is Taro's multivariable Ackermann function * { } is BEAF * \(f_\alpha(n)\) is Fast-growing hierarchy * \(D\) is function in loader.c * \(\Sigma\) is Rado's sigma function * \(R_\alpha(n)\) is Rayo hierarchy Calculation When \(47 \le x \le 71\), \(M(x) = H(^{x/20}2,2)\), where \(H\) is H function. \begin{eqnarray*} H(0,2) &=& H_{0}(2) = 2 \\ H(1,2) &=& H_{1}(2) = 3 \\ H(2,2) &=& H_{\omega}(2) = 4 \\ H(3,2) &=& H_{\omega+1}(2) = 6 \\ H(4,2) &=& H_{\omega^\omega}(2) = H_{\omega+2}(2) = 8 \\ H(5,2) &=& H_{\omega^\omega+1}(2) = H_{\omega^\omega}(3) = H_{\omega^2 2 + \omega 2 + 3}(3) = f_2^2(24) \approx 10^{10^8} \\ H(6,2) &=& H_{\omega^\omega+\omega}(2) = H_{\omega^\omega+2}(2) = H_{\omega^\omega}(4) \approx 5↑↑↑5 \\ H(7,2) &=& H_{\omega^\omega+\omega+1}(2) = H_{\omega^\omega+\omega}(3) = H_{\omega^\omega}(6) \approx 7↑^{5}7 \\ H(8,2) &=& H_{\omega^{\omega+1}}(2) = H_{\omega^\omega・2}(2) = H_{\omega^\omega}^2(2) = H_{\omega^\omega}(8) \approx 9↑^{7}9 \\ H(9,2) &=& H_{\omega^{\omega+1}+1}(2) = H_{\omega^{\omega+1}}(3) = f_{\omega+1}(3) = f_{\omega}^3(3) \\ &=& f_{\omega}^2(H_{\omega^2 2 + \omega 2 +3}(3)) \approx f_{\omega}^2(10^{10^8}) > f_{\omega}^2(10^{10}) \approx f_{\omega}(10 \rightarrow 10 \rightarrow 2 \rightarrow 2)\\ &\approx& 10 \rightarrow 10 \rightarrow 3 \rightarrow 2 \\ H(10,2) &=& H_{\omega^{\omega+1}+\omega}(2) = H_{\omega^{\omega+1}}(4) = f_{\omega+1}(4) \approx 10 \rightarrow 10 \rightarrow 4 \rightarrow 2 \\ H(11,2) &=& H_{\omega^{\omega+1}+\omega+1}(2) = H_{\omega^{\omega+1}+\omega}(3) = f_{\omega+1}(6) \approx 10 \rightarrow 10 \rightarrow 6 \rightarrow 2 \\ H(12,2) &=& H_{\omega^{\omega+1}+\omega^\omega}(2) = H_{\omega^{\omega+1}}(H_{\omega^\omega}(2)) = f_{\omega+1}(8) \approx 10 \rightarrow 10 \rightarrow 8 \rightarrow 2 \\ H(13,2) &=& H_{\omega^{\omega+1}+\omega^\omega+1}(2) = H_{\omega^{\omega+1}}(H_{\omega^\omega+1}(2)) \approx A(1,1,10^{10^8}) \\ H(14,2) &=& H_{\omega^{\omega+1}+\omega^\omega+\omega}(2) = H_{\omega^{\omega+1}}(H_{\omega^\omega+\omega}(2)) \approx A(1,1,5↑↑↑5) \\ H(15,2) &=& H_{\omega^{\omega+1}+\omega^\omega+\omega+1}(2) = H_{\omega^{\omega+1}}(H_{\omega^\omega+\omega+1}(2)) \approx A(1,1,7 \uparrow^5 7) \\ H(16,2) &=& H_{\omega^{\omega^\omega}}(2) = f_{\omega^\omega}(2) = f_{\omega+2}(2) = f_{\omega+1}^2(2) \\ &=& f_{\omega+1}(f_{\omega}^2(2)) = f_{\omega+1}(f_{\omega}(8)) \\ &\approx& A(1,1,3 \rightarrow 8 \rightarrow 7) \\ H(17,2) &=& H_{\omega^{\omega^\omega}+1}(2) = f_{\omega^\omega}(3) \approx A(1,0,0,0,3) = A(2,2,3,3) \\ H(18,2) &=& H_{\omega^{\omega^\omega}+\omega}(2) = f_{\omega^\omega}(4) \approx A(1,0,0,0,0,4) = A(3,3,3,4,4) \\ H(19,2) &=& H_{\omega^{\omega^\omega}+\omega+1}(2) = f_{\omega^\omega}(6) \approx A(1,0,0,0,0,0,0,6) = A(5,5,5,5,5,6,6) \\ H(20,2) &=& H_{\omega^{\omega^\omega}+\omega^\omega}(2) = f_{\omega^\omega}(8) \approx A(1,0,0,0,0,0,0,0,0,8) = A(7,7,7,7,7,7,7,8,8) \\ H(21,2) &=& H_{\omega^{\omega^\omega}+\omega^\omega+1}(2) \approx f_{\omega^\omega}(10^{10^8}) \\ \end{eqnarray*} Interpolation between \(H(5,2)\) and \(H(6,2)\) \begin{eqnarray*} H(5,2) &=& H(27,3) = H_{\omega^\omega}(3) = H_{\omega^3}(3) = H_{\omega^2 3}(3) = H_{\omega^2 2 +\omega 2 + 3}(3) \\ &=& H_{\omega^2 2 +\omega 2 + 2}(4) = H(4^2*2 + 4*2 + 2,4) = H(32+8+2,4) = H(42,4) \\ H(6,2) &=& H(28,3) = H_{\omega^\omega + 1}(3) \\ &=& H_{\omega^\omega}(4) = H(256,4) \\ \end{eqnarray*} Therefore, \(H(5+r,2) = H(42+(256-42)r,2) = H(42+214r)\) \begin{eqnarray*} M(47) &=& H(^{2.35}2,2) = H(2^{2^{2^{0.35}}},2) = H(5.34893, 2) \\ &=& H(42+214 \times 0.34893,4) = H(116.67178, 4) \\ &>& H(116,4) = H_{\omega^3+\omega^2*3+\omega}(4) = f_3(f_2^3(f_1(4))) = f_3(f_2^3(8)) \\ &>& f_3(2^{2^{2059}}) > 10 \uparrow \uparrow 10^{10^{619}} \\ M(48) &=& H(5.64045, 2) = H(42+214 \times 0.64045,4) = H(179.05579) \\ &>& H(179, 4) = f_3^2(f_2^3(7))) > f_3^2(2^{2^{905}}) > 4 \uparrow \uparrow 4 \uparrow \uparrow 10^{10^{271}} \\ &>& 4 \uparrow \uparrow 4 \uparrow \uparrow 4 \uparrow \uparrow 4 = 4 \uparrow \uparrow \uparrow 4 = \{4, 4, 3\} \\ M(49) &=& H(5.96955, 2) = H(42+214 \times 0.96955,4) = H(249.48282) \\ &>& H(249, 4) = f_3^3(f_2^3(f_1^2(5))) = f_3^3(f_2^3(20)) > f_3^3(2^{2^{20971544}}) \\ &>& 4 \uparrow \uparrow 4 \uparrow \uparrow 4 \uparrow \uparrow 10^{10^{6313063}} > 4 \uparrow \uparrow 4 \uparrow \uparrow 4 \uparrow \uparrow 4 \uparrow \uparrow 4 \\ &=& 4 \uparrow \uparrow \uparrow 5 = \{4, 5, 3\} \\ \end{eqnarray*} Interpolation between \(H(6,2)\) and \(H(7,2)\). Since \(H(6,2) = H(28,3)\) and \(H(7,2) = H(30,3)\), \(H(6.5,2) = H(29,3)\). \begin{eqnarray*} H(6,2) &=& H(28,3) = H(256,4) = H(5^3 \times 3 + 5^2 \times 3 + 5 \times 3 + 3,5) = H(468,5) \\ H(6.5,2) &=& H(29,3) = H(257,4) = H(3125,5) = H(6220,6) \\ H(7,2) &=& H(30,3) = H(258,4) = H(3126,5)= H(46656,6) \\ \end{eqnarray*} Therefore, \begin{eqnarray*} M(50) &=& H(6.3429,2) = H(28.6858,3) = H(256.6858,4) \\ &=& H(468+2657 \times 0.6858, 5) > H(2290,5) \\ &=& f_4^3(f_3^3(f_2(f_1^3(5)))) = f_4^3(f_3^3(f_2(40))) > f_4^3(f_3^3(2^{45})) \\ &\approx& f_4^3(3 \uparrow \uparrow 3 \uparrow \uparrow 3 \uparrow \uparrow 3 \uparrow \uparrow 3) = f_4^3(3 \uparrow^3 5) > f_4^3(4 \uparrow^3 4) \\ &=& 4 \uparrow^3 4 \uparrow^3 4 \uparrow^3 4 \uparrow^3 4 = \{4, 5, 4\} \\ M(51) &=& H(6.76873,2) = H(29.53747,3) = H(3125.53747,5) \\ &=& H(6220+40436 \times 0.53747, 6) > H(27952,6) = f_5^3(f_4^3(f_3^3(f_2^2(f_1^2(10))))) \\ &=& f_5^3(f_4^3(f_3^3(f_2^2(40)))) > f_5^3(f_4^3(f_3^3(2^{45}))) > f_5^3(f_4^3(4)) \\ &\approx& f_5^3(4 \uparrow^4 4) = 4 \uparrow^4 4 \uparrow^4 4 \uparrow^4 4 \uparrow^4 4 = \{4, 5, 5\} \\ M(51) &<& H(27953,6) = f_5^3(f_4^3(f_3^3(f_2^2(f_1^2(11))))) = f_5^3(f_4^3(f_3^3(f_2^2(44)))) \\ &<& f_5^3(f_4^3(f_3^3(2^{50}))) < f_5^3(f_4^3(f_3^3(5 \uparrow^2 5))) \approx f_5^3(f_4^3(5 \uparrow^3 5)) \approx f_5^3(5 \uparrow^4 5) \\ &\approx& \{5, 5, 5\} \end{eqnarray*} Interpolation between \(H(7,2)\) and \(H(8,2)\). Since \(H(7,2) = H(30,3) = H(258,4)\) and \(H(8,2) = H(31,3) = H(260,4)\), \(H(7.5,2) = H(259,4)\). \begin{eqnarray*} H(7,2) &=& H(258,4) = H(6^6,6) = H(98040,7) \\ H(7.5,2) &=& H(259,4) = H(7^7,7) = H(823543,7) = H(1797558,8) \\ H(8,2) &=& H(260,4) = H(8^8,8) = H(16777216,8) \\ \end{eqnarray*} Therefore, \begin{eqnarray*} M(52) &=& H(7.25718,2) = H(258.51435,4) = H(46656.51435,6) > H(471203,7) \\ &=& f_6^4(f_3(f_2^5(f_1^2(12)))) > f_6^4(10) \\ &\approx& 10 \uparrow^5 10 \uparrow^5 10 \uparrow^5 10 \uparrow^5 10 = 10 \uparrow^6 5 = 10 \rightarrow 5 \rightarrow 6 \\ M(53) &=& H(7.8209,2) = H(259.6418,4) = H(823543.6418,7) > H(11411690,8) \\ &=& f_7^5(f_6^3(f_5^4(f_4^2(f_2^3(f_1^5(10)))))) > f_7^5(f_6^3(f_5^4(5))) \\ &>& f_7^5(f_6^3(5 \uparrow^3 5 \uparrow^3 5 \uparrow^3 5 \uparrow^3 5)) \\ &=& f_7^5(f_6^3(5 \uparrow^4 5) = f_7^5(5 \uparrow^5 5 \uparrow^5 5 \uparrow^5 5 \uparrow^5 5) \\ &=& f_7^5(5 \uparrow^6 5) = 5 \uparrow^6 5 \uparrow^6 5 \uparrow^6 5 \uparrow^6 5 \uparrow^6 5 \uparrow^6 5 = 5 \uparrow^7 7 = 5 \rightarrow 7 \rightarrow 7 \\ \end{eqnarray*} Interpolation between \(H(8,2)\) and \(H(9,2)\). Since \(H(8,2) = H(31,3)\) and \(H(9,2) = H(81,3)\), \(M(54) = H(8.47588,2) = H(54.794,3)\). As \(H(54,3) = H(298,4)\) and \(H(55,3) = H(512,4)\), \begin{eqnarray*} M(54) &=& H(54.794,3) = H(467.912,4) > H(467,4) \\ &=& H_{\omega^\omega + \omega^3 3 + \omega^2 + 3}(4) = f_\omega(f_3^3(f_2(7))) \\ &=& f_\omega(f_3^3(896)) \approx f_\omega(10 \uparrow^3 4) \\ &\approx& 10 → 10 → (10 → 4 → 3) \\ \end{eqnarray*} Calculation of \(M(55)\) to \(M(57)\). \begin{eqnarray*} M(55) &=& H(9.2424,2) > H(9,2) \approx 10 \rightarrow 10 \rightarrow 3 \rightarrow 2 \\ M(56) &=& H(10.146,2) > H(10,2) \approx 10 \rightarrow 10 \rightarrow 4 \rightarrow 2 \\ M(57) &=& H(11.222,2) > H(11,2) \approx 10 \rightarrow 10 \rightarrow 6 \rightarrow 2 \\ \end{eqnarray*} Calculation of M(58). \begin{eqnarray*} M(58) = H(12.5131,2) \approx A(1,1,H(4.5131,2)) \\ H(4.5131) = H(15.8022,3) \\ \end{eqnarray*} As \(H(15,3) = H(70,8)\) and \(H(16,3) = H(72,8) = H(256,16)\), \(H(15.5,3) = H(71,8) = H(238,16)\), and \begin{eqnarray*} H(15.8022,3) &=& H(248.8791,16) \approx 2^{15} \times 24.8791 \\ &\approx& 8 \times 10^5 \\ \end{eqnarray*} Therefore, \(M(58) \approx A(1,1,8 \times 10^5)\) Calculation of \(M(59)\) and \(M(60)\). \begin{eqnarray*} M(59) &=& H(14.0793,2) > H(14,2) \approx A(1,1,5 \uparrow \uparrow \uparrow 5) \\ M(60) &=& H(16,2) \approx A(1,1,3 \rightarrow 8 \rightarrow 7) \\ \end{eqnarray*} Approximation of \(M(61)\) and \(M(62)\). \begin{eqnarray*} H(17,2) &=& H(3^{3^3},3) \approx H(4^{4^2*2+4*2+2}*2) \approx H(4 \times 10^{25},4) \\ H(18,2) &=& H(3^{3^3}+1,3) = H(4^{4^4},4) \approx H(1.3408 \times 10^{154},4) \\ A(3,3,3,3,4) &\approx& H_{\omega^{\omega^3 3 + \omega^2 3 + \omega 3 + 3}}(4) \\ &\approx& H(3.352 \times 10^{153},4) \approx H(17.25,2) \\ M(61) &=& H(17.6255,2) > H(17+1/4) \approx A(3,3,3,3,4) \\ M(62) &=& H(19.6042,2) > H(19+1/8) \approx A(7,7,7,7,7,7,7,7,8) \\ \end{eqnarray*} Calculation of \(M(63)\) to \(M(65)\). \begin{eqnarray*} M(63) &=& H(22.045,2) = f_{\omega^{\omega}}(H(6.045,2)) \approx f_{\omega^{\omega}}(H(206,5)) \approx f_{\omega^{\omega}}(4 \uparrow \uparrow \uparrow 3) \\ M(64) &=& H(25.1015,2) = f_{\omega^{\omega}}(H(9.1015,2)) \approx f_{\omega^{\omega}}(3 \rightarrow 3 \rightarrow 3 \rightarrow 2) \\ M(65) &=& H(28.9944,2) = f_{\omega^{\omega}}(H(12.9944,2)) \approx f_{\omega^{\omega}}(A(1,1,10^{10^8}) \\ \end{eqnarray*} Calculation of \(M(66) = H(34.0487,2)\). \begin{eqnarray*} H(34,2) &=& f_{\omega^{\omega}+1}(4) = H(4^{4^4+1},4) \approx H(5^{5^5} \cdot 4,5) \\ H(35,2) &=& H(4^{4^4+1}+2,4) = f_{\omega^{\omega}+1}(6) \\ H(34.5,2) &=& H(4^{4^4+1}+1,4) = H(5^{5^5+1},5) = f_{\omega^{\omega}+1}(5) \\ M(66) &\approx& H(34.05,2) \approx H(5^{5^5} \cdot 4 + 5^{5^5} \cdot 0.1,5) > H(5^{5^5} \cdot 4 + 5^{5^5-2},5) \\ \end{eqnarray*} Therefore, \(f_{\omega^{\omega}+1}(4) < M(66) < f_{\omega^{\omega}+1}(5)\) and \(M(66)\) is better approximated as \(f_{\omega^{\omega}+1}(5)\). Calculation of \(M(67)\). Since \(H(8,2) = H(31,3)\) and \(H(9,2) = H(81,3)\), \begin{eqnarray*} M(67) &=& H(40.7558,2) = f_{\omega^{\omega}+1}(H(8.7558,2)) = f_{\omega^{\omega}+1}(H(68.7901,3)) \\ &=& f_{\omega^{\omega}+1}(f_{\omega}^2(H(14.7901,3))) \approx f_{\omega^{\omega}+1}(f_{\omega}^2(H_{\omega^2 + \omega + 2}(3))) \\ &\approx& f_{\omega^{\omega}+1}(f_{\omega}^2(10^4)) \\ &\approx& f_{\omega^{\omega}+1}(10 \rightarrow 4 \rightarrow 3 \rightarrow 2) \\ \end{eqnarray*} Calculation of \(M(68)\) to \(M(71)\). \begin{eqnarray*} M(68) &=& H(49.882,2) = f_{\omega^{\omega}+1}(H(17.882,2)) \approx f_{\omega^{\omega}+1}(A(3,3,3,3,4)) \\ M(69) &=& H(62.6632,2) = f_{\omega^{\omega}+1}(f_{\omega^{\omega}}(H(14.6632,2))) \\ &\approx& f_{\omega^{\omega}+1}(f_{\omega^{\omega}}(A(1,1,5 \uparrow \uparrow \uparrow5))) \\ M(70) &=& H(81.1718,2) = f_{\omega^{\omega}+\omega}(H(17.1718,2)) \\ &\approx& f_{\omega^{\omega}+\omega}(A(3,3,3,2,4)) \\ M(71) &=& H(109.041,2) = f_{\omega^{\omega}+\omega}(H(45.041,2)) \\ &=& f_{\omega^{\omega}+\omega}(f_{\omega^{\omega}+1}(H(13.041,2)) \\ &\approx& f_{\omega^{\omega}+\omega}(f_{\omega^{\omega}+1}(A(1,1,10^{10^8}))) \\ \end{eqnarray*} Sources Category:Blog posts